Anindya wrote:

> from what I can make out from the proof of the strictification theorem it doesn't actually collapse all isomorphic objects into one object – it just identifies, e.g.. \$$A\otimes (B\otimes C)\$$ and \$$(A\otimes B)\otimes C\$$ for all objects \$$A, B, C\$$. So the strict monoidal category isn't necessarily skeletal.

The strict monoidal category is indeed not necessarily skeletal, as you can see from my last comment.

But your first sentence is hovering on the brink between truth and falsehood, as you can also see from my last comment. If in the monoidal category \$$\mathcal{C}\$$ we have two unequal objects

$$X = A\otimes (B\otimes C)$$

and

$$Y = ((A\otimes B)\otimes C ,$$

there will also be two unequal objects \$$(X) \$$ and \$$(Y) \$$ in the strictification \$$\mathrm{str}(\mathcal{C})\$$. These are one-element lists.

In short, we are not identifying any objects that were previously unequal! Instead, we are throwing in a lot _more_ objects, and creating a _new_ tensor product that is strictly associative.